Some Basic Concepts in Chemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

\[\underset{\begin{smallmatrix} 1\,mole \\ \frac{1}{2}\,mole \end{smallmatrix}}{\mathop{Mg}}\,+2HCl\to MgC{{l}_{2}}+\underset{\begin{smallmatrix} 1\,mole \\ \frac{1}{2}\,mole\,\left( 12g\,\,\text{of}\,Mg=\frac{1}{2}\,mol \right) \end{smallmatrix}}{\mathop{{{H}_{2}}\uparrow }}\,\]

$${\text{Molality}}$$   $$ = \frac{{{\text{Wt}}{\text{. of solute}}}}{{{\text{Mol}}{\text{. wt}}{\text{. of solute}}}} \times \frac{{100}}{{{\text{Wt}}{\text{. of solvent}}}}$$
$$\eqalign{ & m = \frac{{10}}{{40}} \times \frac{{1000}}{{100}} \cr & \,\,\,\,\,\, = 2.5\,m \cr} $$

\[\begin{align} & \underset{\begin{smallmatrix} 2\times 18 \\ =36\,g \end{smallmatrix}}{\mathop{2{{H}_{2}}O}}\,\to 2{{H}_{2}}+{{O}_{2}} \\ & 36\,g\,\,\text{of water produces}\,\,1\,mole\,\,\text{of}\,\,{{O}_{2}}\,\,\text{gas}\text{.} \\ & 180\,g\,\,\text{of water will produce}\,\,\frac{180}{36}=5\,\,moles\,\,\text{of}\,\,{{O}_{2}}\,\,\text{gas}. \\ \end{align}\]

$$\eqalign{ & \mathop {Ba{{\left( {OH} \right)}_2}}\limits_{1\,mol} + C{O_2} \to \mathop {BaC{O_3}}\limits_{1\,mol} + {H_2}O \cr & 1\,mol\,Ba{\left( {OH} \right)_2} = 1\,mol\,BaC{O_3} \cr} $$
$$\therefore \,\,0.205\,mol\,Ba{\left( {OH} \right)_2} \equiv $$       $$0.205\,mol\,BaC{O_3}$$
$$Wt.\,{\text{of substance}} = $$     $${\text{No}}{\text{. of}}\,moles \times {\text{Molecular}}\,mass$$
$$\eqalign{ & = 0.205 \times 197.3 \cr & = 40.5\,g \cr} $$

$$\eqalign{ & Ti + {O_2} \to T{i_{1.44}}O \cr & \frac{{1.44}}{{48}}mole\,\frac{x}{{48\left( {1.44} \right) + 16}}mole \cr & \therefore \,\,\frac{{1.44}}{{48}} = \frac{{1.44x}}{{48\left( {1.44} \right) + 16}} \cr & x = 1.77\,g \cr} $$

TIPS/Formulae :
(i) Write balance chemical equation for given change.
(ii) Identify most electronegative element and find its oxidation state.
$$Ba{O_2} + {H_2}S{O_4} \to BaS{O_4} + {H_2}{O_2}$$
Oxygen is the most electronegative element in the reaction and has the oxidation states of $$ - 1\left( {{\text{in}}\,{H_2}{O_2}} \right)\,{\text{and}}\, - 2\left( {{\text{in}}\,BaS{O_4}} \right).\,\,{\text{In}}\,{H_2}{O_{2,}}$$         peroxo ion is present.

TIPS/Formulae :
(i) Volume of substance changes with temperature and mass is not effected by change in temperature.
(ii) Find expression which does not have volume term in it.
(A) Molarity — Moles of solute/volume of solution in $$L$$
(B) Normality — gm equivalents of solute/volume of solution in $$L.$$
(C) Formality—gm formula wt./volume of solution in $$L.$$
(D) Molality — Moles of solute/mass of solvent in $$kg$$
$$\because $$ Molality does not involve volume term.
$$\therefore $$  It is independent of temperature.

$$\eqalign{ & 5CO + {I_2}{O_5} \to 5C{O_2} + {I_2} \cr & {\text{Moles of}}\,{I_2}{O_5} = \frac{{25.4}}{{254}} \cr & = 0.01 \equiv 0.05\,{\text{moles of}}\,\,CO \cr & {\text{Weight of}}\,\,CO = 0.05 \times 28 = 1.4\,g; \cr & {\text{Weight of}}\,\,C{O_2} = 2 - 1.4 = 0.6\,g \cr & {\text{Hence}}\,\,\% \,\,{\text{of}}\,\,C{O_2} = \frac{{0.6}}{2} \times 100 = 30\% \cr} $$

The $$H : O$$  ratio in water is fixed, irrespective of its source. Hence it is law of constant composition.

$$4C{a_5}\mathop {{{\left( {P{O_4}} \right)}_3}}\limits_{0.1} F + \mathop {18Si{O_2}}\limits_{0.36}^{L.R} + \mathop {30C}\limits_{0.9} \to $$       $$3{P_4} + 2Ca{F_2} + 18CaSi{O_3} + 30CO$$
$$\eqalign{ & {\text{18 moles of}}\,Si{O_2}\,{\text{gives 3 moles of }}{P_4} \cr & 0.36\,moles\,{\text{of}}\,Si{O_2}\,{\text{will give}} \cr & \frac{3}{{18}} \times 0.36 = 0.06\,mole \cr} $$